Advanced Math Solutions - Series Convergence Calculator, Alternating Series Test

Last blog post, we discussed how to determine if an infinite p-series converges using the p-series test. In this blog post, we will discuss how to determine if an infinite alternating series converges using the alternating series test.

An alternating series is a series in the form ∑_{n=0}^∞(-1)^n∙a_n or ∑_{n=0}^∞(-1)^{n-1}∙a_n, where a_n>0 for all n. As you can see, the alternating series got its name from its terms that alternate between positive and negative values.

In some problems, you may come across alternating sign expressions that aren’t listed above. These various sign expressions can be simplified to equal the ones listed above. For example:

                  (-1)^{n+1}=(-1)^{n-1}∙(-1)^2=(-1)^{n-1}∙1=(-1)^{n-1}

Now that we know what an alternating series is, let’s discuss how to determine if the series converges, using the alternating series test.

Alternating Series Test:

An alternating series converges if all of the following conditions are met:
1. a_n>0 for all n

• a_n is positive

2. a_n>a_(n+1)  for all n≥N,where N is some integer

• a_n is always decreasing

3. lim_{n→∞} a_n=0

If an alternating series fails to meet one of the conditions, it doesn’t mean the series diverges. There are other tests that can be used to determine divergence.

Let’s see some examples of how to use the alternating series test to help you better understand.

First example (click here):

                                            ∑_{n=1}^∞\frac{(-1)^{n+1}}{\sqrt{n+1}}
1. Is a_n>0 for all n

                  ∑_{n=1}^∞\frac{(-1)^{n+1}}{\sqrt{n+1}}=∑_{n=1}^∞(-1)^{n+1}∙\frac{1}{\sqrt{n+1}}

                                                 a_n=\frac{1}{\sqrt{n+1}}>0 for all n

2. Is a_n>a_(n+1)  for all n≥N,where N is some integer?

                                   ∑_{n=1}^∞\frac{1}{\sqrt{n+1}}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+ ...
                                         \frac{1}{\sqrt{2}}>\frac{1}{\sqrt{3}}  and  \frac{1}{\sqrt{3}}>\frac{1}{\sqrt{4}}  and so on

                                                  a_n>a_(n+1)  for all n≥1

3. Is lim_{n→∞} a_n=0?

      Let’s use this property to determine the limit of a_n:

                                   lim_{n→∞}  \frac{1}{\sqrt{n+1}}=\frac{lim_{n→∞} 1}{lim_{n→∞} \sqrt{n+1}}=\frac{1}{∞}=0

4. Are all three conditions met?

                                                   Yes, the series converges.

Next example (click here):

                                                ∑_{n=1}^∞\frac{(-1)^n (n)}{n^2+2}

1. Is a_n>0 for all n?

                                                     a_n=\frac{n}{n^2+2}>0 for all n

2. Is a_n>a_{n+1}  for all n≥N,where N is some integer?

                   ∑_{n=1}^∞\frac{n}{n^2+2}=\frac{1}{3}+\frac{2}{6}+\frac{3}{11}+\frac{4}{18}  ...=\frac{1}{3}+\frac{1}{3}+\frac{3}{11}+\frac{2}{9}  ...

                                        \frac{1}{3}=\frac{1}{3},   \frac{1}{3}>\frac{3}{11},   \frac{3}{11}>\frac{2}{9}, and so on

                                                     a_n>a_(n+1)  for all n≥2

3. Is lim_{n→∞} a_n=0?

                                              lim_{n→∞}  \frac{n}{n^2+1}=lim_{n→∞}  \frac{\frac{1}{n}}{1+\frac{1}{n^2}}=\frac{lim_{n→∞}  \frac{1}{n}}{lim_{n→∞} 1+\frac{1}{n^2}}=  \frac{0}{1+0}=0

4. Are all three conditions met?

                                                     Yes, the series converges.

Last example (click here):

                                                 ∑_{n=2}^∞\frac{(-1)^{n+2}}{ln⁡(n)}

As discussed in the beginning, we can see a different alternating sign expression that wasn’t listed. However, we can simplify this expression to equal one that was listed.

                                    (-1)^{n+2}=(-1)^n*(-1)^2=(-1)^n

1. Is a_n>0 for all n?

                                                    a_n=\frac{1}{ln⁡(n)}>0 for all n

2. Is a_n>a_{n+1}  for all n≥N,where N is some integer?

                              ∑_{n=2}^∞\frac{1}{ln⁡(n)}=\frac{1}{ln⁡(2)}+\frac{1}{ln⁡(3)}+\frac{1}{ln⁡(4)}  ...

                                 \frac{1}{ln⁡(2)}>\frac{1}{ln⁡(3)},   \frac{1}{ln⁡(3)}>\frac{1}{ln⁡(4)},    and so on

                                                   a_n>a_{n+1}  for all n≥2

3. Is lim_{n→∞} a_n=0?

                                 lim_{n→∞}  \frac{1}{ln⁡(n)}=\frac{lim_{n→∞} 1}{lim_{n→∞} ln⁡(n)}=\frac{1}{∞}=0

4. Are all three conditions met?

                                                     Yes, the series converges.

The alternating series test has a lot of parts to it, but as long as you remember the three conditions, you’ll be able to master this topic! For more help or practice on the alternating series test, check out Symbolab’s Practice. Next blog post, I’ll go over the convergence test for telescoping series.

Until next time,

Leah